/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 To pull a \(50 \mathrm{~kg}\) cr... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 58

To pull a \(50 \mathrm{~kg}\) crate across a horizontal frictionless floor, a worker applies a force of \(210 \mathrm{~N},\) directed \(20^{\circ}\) above the horizontal. As the crate moves \(3.0 \mathrm{~m},\) what work is done on the crate by (a) the worker's force, (b) the gravitational force, and (c) the normal force? (d) What is the total work?

Short Answer

Expert verified
The total work done is approximately 592.58 J.

Step by step solution

01

Understanding Work Done by the Worker鈥檚 Force

The work done by a force when moving an object is given by the formula: \( W = Fd \cos \theta \), where \( F \) is the force applied, \( d \) is the distance moved, and \( \theta \) is the angle between the force and the direction of motion. Here, \( F = 210 \text{ N} \), \( d = 3.0 \text{ m} \), and \( \theta = 20^\circ \). Substitute these values to calculate the work: \[ W = 210 \times 3.0 \times \cos 20^\circ \approx 592.58 \text{ J} \].
02

Calculating Work Done by the Gravitational Force

The gravitational force acts vertically downward, while the crate moves horizontally, so the angle between the gravitational force and the direction of motion is \(90^\circ\). The work done is \( W = F_{gravity} \cdot d \cdot \cos 90^\circ = 0 \text{ J} \), because \(\cos 90^\circ = 0\).
03

Finding Work Done by the Normal Force

Similar to the gravitational force, the normal force is perpendicular to the direction of motion, making the angle \(90^\circ\) again. Hence, the work done by the normal force is also \(0 \text{ J}\), using \( W = F_{normal} \cdot d \cdot \cos 90^\circ = 0 \text{ J}\).
04

Calculating the Total Work Done

The total work done on the crate is the sum of all the work done by each force acting on it. Therefore, the total work is \[ W_{total} = W_{worker} + W_{gravity} + W_{normal} = 592.58 \text{ J} + 0 \text{ J} + 0 \text{ J} = 592.58 \text{ J} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work done by a force
When calculating work done by a force, we need to consider the magnitude of the force, the distance over which the force is applied, and the angle between the force and the direction of motion.
  • Formula: The standard formula is \( W = Fd \cos \theta \), where \( W \) is the work done, \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and motion.
  • Angle Importance: If the angle \( \theta \) is 0 degrees, the force is fully aligned with the direction of motion, leading to maximum work done. At 90 degrees, no work is done because the force is perpendicular to the motion.
In the scenario with the crate, the worker applies a force at a 20-degree angle to the horizontal.
Apply the formula with \( F = 210 \text{ N} \), \( d = 3.0 \text{ m} \), and \( \theta = 20^\circ \) to find \( W = 592.58 \text{ J} \), showing how force can effectively move an object over a distance, even if applied at an angle.
Gravitational force
The gravitational force is a fundamental force that acts downwards towards the center of the Earth. It always acts perpendicular to any horizontal motion. When an object moves horizontally on a frictionless surface, the work done by gravitational force is effectively zero.
  • Force Direction: Since gravity acts vertically, when an object moves horizontally, the angle between gravity and the direction of motion is 90 degrees.
  • Work Done: The equation \( W = F_{gravity} \cdot d \cdot \cos \theta \) gives zero work because \( \cos 90^\circ = 0 \).
Thus, in the case of the crate being pulled across the floor, no work is performed by gravity because its force doesn't contribute to the movement in the direction the crate is moving.
Normal force
Normal force is the force that acts perpendicular to the contact surface. For an object resting on a horizontal surface, the normal force balances the gravitational force.
  • Perpendicular Forces: Since the normal force is perpendicular to the direction of horizontal movement, it also forms a 90-degree angle with the direction of motion.
  • Work Done: Just like gravity, the work done by the normal force is zero because \( W = F_{normal} \cdot d \cdot \cos 90^\circ = 0 \).
In the case of the crate, the normal force only supports the weight of the crate and doesn't add any work in moving the crate horizontally.
Understanding this highlights why only forces with a component in the direction of motion do actual work in moving objects.

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Most popular questions from this chapter

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of \(10.0 \mathrm{~m}:\) (a) the initially stationary spelunker is accelerated to a speed of \(5.00 \mathrm{~m} / \mathrm{s} ;\) (b) he is then lifted at the constant speed of \(5.00 \mathrm{~m} / \mathrm{s}\) (c) finally he is decelerated to zero speed. How much work is done on the \(80.0 \mathrm{~kg}\) rescuee by the force lifting him during each stage?

A \(4.00 \mathrm{~kg}\) block is pulled up a frictionless inclined plane by a \(50.0 \mathrm{~N}\) force that is parallel to the plane, starting from rest. The normal force on the block from the plane has magnitude \(13.41 \mathrm{~N}\). What is the block's speed when its displacement up the ramp is \(3.00 \mathrm{~m} ?\)

A force of \(5.0 \mathrm{~N}\) acts on a \(15 \mathrm{~kg}\) body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

A \(12.0 \mathrm{~N}\) force with a fixed orientation does work on a particle as the particle moves through the three-dimensional displacement \(\vec{d}=(2.00 \hat{i}-4.00 \mathrm{j}+3.00 \mathrm{k}) \mathrm{m} .\) What is the angle between the force and the displacement if the change in the particle's kinetic energy is (a) \(+30.0 \mathrm{~J}\) and (b) \(-30.0 \mathrm{~J} ?\)

A 100 kg block is pulled at a constant speed of \(5.0 \mathrm{~m} / \mathrm{s}\) across a horizontal floor by an applied force of \(122 \mathrm{~N}\) directed \(37^{\circ}\) above the horizontal. What is the rate at which the force does work on the block?
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